If [x] is the greatest integer in x, then what is lim [x+1] as x approaches -1?

Question

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Answer 1

The limit doesn't exist.
The limit exists only when the limit from the left and the limit from the right approach the same value.

The limit as you approach from the left is -1.
The limit as you approach from the right is 0.

Answer 2

0 x+1=-1+1=0

Answer 3

The minimize would not exist because you get diverse habit from left and proper: From the left, we've x = -a million.a million, -a million.01, -a million.001, etc. accordingly you get we've x = -a million - delta --> a million + x = -delta --> i.e. -0.a million, -0.01, -0.001, etc. The minimize will be -a million, because you're not from now on truly at 0 yet. on the different hand, if we mind-set from the right we get x = -.9, -.ninety 9, -.999, etc. accordingly we've x = -a million + delta --> a million + x = +delta --> i.e. 0.a million, 0.01, 0.001, etc. The minimize the following will be 0 (back because it is the biggest integer). So the minimize would not exist because of diverse left-proper habit. Edit: btw, the very undeniable reality that the minimize would not exist @ x = -a million, yet the cost is properly defined: [-a million +a million] = [0] = 0, signifies that this function is discontinuous at x = -a million. And more desirable oftentimes, is discontinuous in any respect integer values of x.

Answer 4

Undefined, because the limit from the right is 0, but the limit from the left is -1.

Answer 5

As x becomes increasingly closer to -1 the limit of (x+1) is zero.